Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 200002;
const int INF = 1e9;
int n, sy, a, b, ps[MAX_N];
pair<int, int> fence[MAX_N];
void enter() {
  cin >> sy >> a >> b >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> fence[i].first >> fence[i].second;
  }
}
void init() {
  sort(fence + 1, fence + n + 1);
  for (int i = 1; i <= n; ++i) {
    ps[i] = ps[i - 1] + fence[i].second - fence[i].first;
  }
}
int bisect_left(int x, int y, double x2) {
  int l = 1, r = n, mid = (l + r) / 2;
  while (l != mid && mid != r) {
    if ((double)fence[mid].second >= x2)
      r = mid;
    else
      l = mid;
    mid = (l + r) / 2;
  }
  for (int i = l; i <= r; ++i)
    if ((double)fence[i].second >= x2) return i;
  return n + 1;
}
int bisect_right(int x, int y, double x2) {
  int l = 1, r = n, mid = (l + r) / 2;
  while (l != mid && mid != r) {
    if ((double)fence[mid].first <= x2)
      l = mid;
    else
      r = mid;
    mid = (l + r) / 2;
  }
  for (int i = r; i >= l; --i)
    if ((double)fence[i].first <= x2) return i;
  return 0;
}
double find_common_point(double x1, double _y1_, double x2, double _y2_) {
  return (x2 * _y1_ - x1 * _y2_) / (_y1_ - _y2_);
}
double intersect(double l, double r, double u, double v) {
  return max(0.0, min(r, v) - max(l, u));
}
double dist(double x1, double _y1_, double x2, double _y2_) {
  return sqrt((x1 - x2) * (x1 - x2) + (_y1_ - _y2_) * (_y1_ - _y2_));
}
void solve() {
  int nQuery;
  cin >> nQuery;
  cout << setiosflags(ios::fixed) << setprecision(10);
  while (nQuery--) {
    int x, y;
    cin >> x >> y;
    double x1 = find_common_point(x, y, a, sy);
    double x2 = find_common_point(x, y, b, sy);
    int l = bisect_left(x, y, x1), r = bisect_right(x, y, x2);
    double f = 0;
    if (l <= r) {
      f = (ps[r] - ps[l - 1]);
      f -= intersect(fence[l].first, fence[l].second, 1, x1);
      f -= intersect(fence[r].first, fence[r].second, x2, INF);
    }
    double res = f * dist(x, y, a, sy) / dist(x, y, x1, 0);
    cout << res << '\n';
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  enter();
  init();
  solve();
}